convergent series for $\sum_{n=1}^{\infty}\frac{n!}{n^n}$

Question

Help me please , I am not able to solve this problem.I have tried in many ways to figure out such as Ration test , Integral test , Comparison test , Limit Comparison Test , Root Test but i can't find the way out . This is my first question and i'm not good at English. If there is something wrong or you are not comfortable with my language usage I'm so sorry.

You may find https://math.stackexchange.com/questions/354487/does-the-series-sum-n-nn-converge-or-diverge helpful. — Minus One-Twelfth, Apr 08 '19 at 11:39

score 0 · Answer 1 · answered Apr 08 '19 at 11:40

The usual approach to factorial-based problems is to use Stirling's approximation $n!\approx\sqrt{2\pi n}n^ne^{-n}$, which shows this series converges provided $\sum_{n\ge 1}\sqrt{2\pi n}e^{-n}$ does, which is the case. If you didn't know this result, another option is to note that $$\frac{\frac{(n+1)!}{(n+1)^{n+1}}}{\frac{n!}{n^n}}=\left(1+\frac{1}{n}\right)^{-n}\approx\frac{1}{e},$$completing the problem by the ratio test.

score 0 · Answer 2 · answered Apr 08 '19 at 11:44

If you write the fraction as

$$\frac1n\frac2n\frac3n\cdots\frac nn,$$

you see that half of the factors are below $\dfrac12$ and the rest below $1$, and you can bound the general term by $\dfrac1{2^{n/2}}$.

convergent series for $\sum_{n=1}^{\infty}\frac{n!}{n^n}$

2 Answers2